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Chapter 16: Problem 2264

The radius of curvature of convex surface of a thin plano-convex lens is $15 \mathrm{~cm}\( and refractive index of its material is \)1.6 .$ The power of the lens will be (A) \(6 \mathrm{D}\) (B) \(5 \mathrm{D}\) (C) \(4 \mathrm{D}\) (D) \(3 \mathrm{D}\)

Short Answer

Expert verified
The power of the lens can be found using the lens maker's formula for a plano-convex lens and the given values of radius of curvature (15 cm) and refractive index (1.6). By calculating the focal length (f = 25 cm) and then the power using the formula Power = 1/focal length, we find that the power of the lens is \(4 \mathrm{D}\). Therefore, the answer is (C) 4D.

Step by step solution

01

Identify the given values

The radius of curvature of the convex surface (R) is 15 cm, and the refractive index (n) is 1.6.
02

Apply the lens maker's formula

The lens maker's formula for a plano-convex lens can be represented as follows:\[ \dfrac{1}{f} = (n - 1) \left[ \dfrac{1}{R_1} - \dfrac{1}{R_2}\right] \]Since the lens is plano-convex, one surface is flat, which means one radius of curvature is infinite (R₂ = ∞). The radius of curvature of the convex side is R₁ = 15 cm. Now, we can insert the given values into the lens maker's formula: \[ \dfrac{1}{f} = (1.6 - 1) \left[ \dfrac{1}{15} - \dfrac{1}{\infty}\right] \]
03

Calculate the focal length

To calculate the focal length from the given values, first simplify the equation: \[ \dfrac{1}{f} = 0.6 \times \dfrac{1}{15} \] \[ \dfrac{1}{f} = \dfrac{1}{25} \] Now, reciprocate both sides to find the focal length (f): \[ f = 25 \mathrm{~cm} \]
04

Calculate the power of the lens

The power of a lens is calculated using the formula Power = 1/focal length, with the focal length expressed in meters: \[ \text{Power} = \dfrac{1}{f} \] \[ \text{Power} = \dfrac{1}{0.25 \mathrm{~m}} \] \[ \text{Power} = \dfrac{1 \times 100}{25} \mathrm{D} \] \[ \text{Power} = 4 \mathrm{D} \]
05

Matching the answer

We have found that the power of the lens is 4D. Thus, the correct answer is (C) 4D.

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